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Interview follow up

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  • L leppie

    TPFKAPB wrote:

    Shouldn't that be 1 repeat and still valid? ie 10 and then 1..

    Nope, that is correct. "repeat implies at least 2 (complete) sequences"

    IronScheme
    ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

    T Offline
    T Offline
    TPFKAPB
    wrote on last edited by
    #11

    Ok so then what about the line below that. 11 => 1 That is only one sequence is it not? Or two sequences but then it should read 11 => 2

    L 1 Reply Last reply
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    • T TPFKAPB

      Ok so then what about the line below that. 11 => 1 That is only one sequence is it not? Or two sequences but then it should read 11 => 2

      L Offline
      L Offline
      leppie
      wrote on last edited by
      #12

      You made a small interpretation mistake (I do it all the time too). The result is the minimum period, not the number of repeats :)

      IronScheme
      ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

      T 1 Reply Last reply
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      • L leppie

        You made a small interpretation mistake (I do it all the time too). The result is the minimum period, not the number of repeats :)

        IronScheme
        ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

        T Offline
        T Offline
        TPFKAPB
        wrote on last edited by
        #13

        Ah yes I remember that part now. Did think it was a bit strange that I thought I could see an error after only a couple of seconds looking at something you had looked at for hours. Thought it must have been me going wrong somewhere and was interested to know where.

        1 Reply Last reply
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        • L leppie

          Was given a problem that was suppose to come from the 'easy' bin. Two hours later I was still not finished... On the plus side, the interviewer attempted the problem at the same time, and also got stuck :) He said the problem must have been miscategorized. He told me to finish at home and to send the solution when done ;p Anyways, after a good nights sleep (with some wack apocalyptic dream I had), I started from scratch and coded a solution in about an hour this morning. The given problem: Binary period (dont bother googling for answers as you will likely just end up getting astronomy results or if you have google foo, a few very mathematical papers mostly related to cryptography) Given a number N larger than 0 (range was given, but does not matter), find the minimum period of repeating bit sequences or -1 if none is found. Examples

          110110110 => 3 (as 110 repeats)
          11011011 => 3 (still 3 and valid as the remaining subset, 11, is in 110)
          110110111 => -1 (no period)
          101 => -1 (no repeats, repeat implies at least 2 sequences)
          10101 => 2 (repeat at least over 2 sequences)
          11 => 1

          10101010101 => 2
          111011011101101110 => 7
          111111111111111110 => -1

          My answer in Scheme can be viewed and run here[^]. Edit: Web host seems to be down now... Edit 2: Seems to be back up again, please only look one at a time ;p Damn you shared hosting. If above link is down, view here[^].

          IronScheme
          ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

          J Offline
          J Offline
          Jonathan Nethercott
          wrote on last edited by
          #14

          This looks like correlation (autocorrelation?) to me. I would look for a solution along the lines of: num xor (num shift left 1) num xor (num shift left 2) etc. repeat until (number of bits / 2) Not a very good bit of pseudo code, but hopefully you get the idea ;) I'm not sure how you'd deal with leading and trailing parts of a sequence, but I'm sure there must be an elegant way of incorporating that...

          Jon CodeWrite

          L 1 Reply Last reply
          0
          • L leppie

            Was given a problem that was suppose to come from the 'easy' bin. Two hours later I was still not finished... On the plus side, the interviewer attempted the problem at the same time, and also got stuck :) He said the problem must have been miscategorized. He told me to finish at home and to send the solution when done ;p Anyways, after a good nights sleep (with some wack apocalyptic dream I had), I started from scratch and coded a solution in about an hour this morning. The given problem: Binary period (dont bother googling for answers as you will likely just end up getting astronomy results or if you have google foo, a few very mathematical papers mostly related to cryptography) Given a number N larger than 0 (range was given, but does not matter), find the minimum period of repeating bit sequences or -1 if none is found. Examples

            110110110 => 3 (as 110 repeats)
            11011011 => 3 (still 3 and valid as the remaining subset, 11, is in 110)
            110110111 => -1 (no period)
            101 => -1 (no repeats, repeat implies at least 2 sequences)
            10101 => 2 (repeat at least over 2 sequences)
            11 => 1

            10101010101 => 2
            111011011101101110 => 7
            111111111111111110 => -1

            My answer in Scheme can be viewed and run here[^]. Edit: Web host seems to be down now... Edit 2: Seems to be back up again, please only look one at a time ;p Damn you shared hosting. If above link is down, view here[^].

            IronScheme
            ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

            S Offline
            S Offline
            Septimus Hedgehog
            wrote on last edited by
            #15

            When will you hear if you've got the job? Best of luck.

            "I do not have to forgive my enemies, I have had them all shot." — Ramón Maria Narváez (1800-68). "I don't need to shoot my enemies, I don't have any." - Me (2012).

            L 1 Reply Last reply
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            • L leppie

              Was given a problem that was suppose to come from the 'easy' bin. Two hours later I was still not finished... On the plus side, the interviewer attempted the problem at the same time, and also got stuck :) He said the problem must have been miscategorized. He told me to finish at home and to send the solution when done ;p Anyways, after a good nights sleep (with some wack apocalyptic dream I had), I started from scratch and coded a solution in about an hour this morning. The given problem: Binary period (dont bother googling for answers as you will likely just end up getting astronomy results or if you have google foo, a few very mathematical papers mostly related to cryptography) Given a number N larger than 0 (range was given, but does not matter), find the minimum period of repeating bit sequences or -1 if none is found. Examples

              110110110 => 3 (as 110 repeats)
              11011011 => 3 (still 3 and valid as the remaining subset, 11, is in 110)
              110110111 => -1 (no period)
              101 => -1 (no repeats, repeat implies at least 2 sequences)
              10101 => 2 (repeat at least over 2 sequences)
              11 => 1

              10101010101 => 2
              111011011101101110 => 7
              111111111111111110 => -1

              My answer in Scheme can be viewed and run here[^]. Edit: Web host seems to be down now... Edit 2: Seems to be back up again, please only look one at a time ;p Damn you shared hosting. If above link is down, view here[^].

              IronScheme
              ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

              M Offline
              M Offline
              Member 2053006
              wrote on last edited by
              #16

              Is the simplistic solution of starting with length 1 and checking each character, then increasing to length 2 etc too processor intensive? e.g. for 11011011 1 1 0 O O - Not 1, so period is not 1. K K 11 01 10 11 OK - Not 11, so period is not 2 110 110 11 All OK - Period is 3. Failing that would a FFT work on this information?

              L 1 Reply Last reply
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              • M Member 2053006

                Is the simplistic solution of starting with length 1 and checking each character, then increasing to length 2 etc too processor intensive? e.g. for 11011011 1 1 0 O O - Not 1, so period is not 1. K K 11 01 10 11 OK - Not 11, so period is not 2 110 110 11 All OK - Period is 3. Failing that would a FFT work on this information?

                L Offline
                L Offline
                leppie
                wrote on last edited by
                #17

                Member 2053006 wrote:

                Failing that would a FFT work on this information?

                That was my first thought, but coding one from scratch with no external help is a bit tough I think :) That is probably the most efficient way to do it.

                IronScheme
                ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

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                • J Jonathan Nethercott

                  This looks like correlation (autocorrelation?) to me. I would look for a solution along the lines of: num xor (num shift left 1) num xor (num shift left 2) etc. repeat until (number of bits / 2) Not a very good bit of pseudo code, but hopefully you get the idea ;) I'm not sure how you'd deal with leading and trailing parts of a sequence, but I'm sure there must be an elegant way of incorporating that...

                  Jon CodeWrite

                  L Offline
                  L Offline
                  leppie
                  wrote on last edited by
                  #18

                  Jon Nethercott wrote:

                  I'm not sure how you'd deal with leading and trailing parts of a sequence, but I'm sure there must be an elegant way of incorporating that...

                  There are no leading parts (thank FSM!). As for the trailing bit, my elegant solution was to do a negative right shift (see line 11, when rs is negative) that fits into the rest (test on line 17). Brain too tired to remember exactly what I did there now ;p

                  IronScheme
                  ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                  J 1 Reply Last reply
                  0
                  • S Septimus Hedgehog

                    When will you hear if you've got the job? Best of luck.

                    "I do not have to forgive my enemies, I have had them all shot." — Ramón Maria Narváez (1800-68). "I don't need to shoot my enemies, I don't have any." - Me (2012).

                    L Offline
                    L Offline
                    leppie
                    wrote on last edited by
                    #19

                    PHS241 wrote:

                    When will you hear if you've got the job?

                    He wants to have another telephonic discussion next week and I guess an offer will be made at that stage, if they so desire.

                    IronScheme
                    ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                    1 Reply Last reply
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                    • L leppie

                      Jon Nethercott wrote:

                      I'm not sure how you'd deal with leading and trailing parts of a sequence, but I'm sure there must be an elegant way of incorporating that...

                      There are no leading parts (thank FSM!). As for the trailing bit, my elegant solution was to do a negative right shift (see line 11, when rs is negative) that fits into the rest (test on line 17). Brain too tired to remember exactly what I did there now ;p

                      IronScheme
                      ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                      J Offline
                      J Offline
                      Jonathan Nethercott
                      wrote on last edited by
                      #20

                      I think this should do it:

                      private int BinaryPeriod(int num)
                      {
                      int result = -1;
                      int numBits = (int)((Math.Log(num) / Math.Log(2))) + 1;

                      for (int i = 1; i <= numBits / 2 && result == -1; i++)
                      {
                          if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)
                              result = i;
                      }
                      return result;
                      

                      }

                      I've done it in C#, but hopefully it should be fairly generic code. The important line is:

                      if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)

                      which strips the top i bits and XORs that with num right shifted. If the result is 0 then we have found an autocorrelation.

                      Jon CodeWrite

                      L 2 Replies Last reply
                      0
                      • J Jonathan Nethercott

                        I think this should do it:

                        private int BinaryPeriod(int num)
                        {
                        int result = -1;
                        int numBits = (int)((Math.Log(num) / Math.Log(2))) + 1;

                        for (int i = 1; i <= numBits / 2 && result == -1; i++)
                        {
                            if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)
                                result = i;
                        }
                        return result;
                        

                        }

                        I've done it in C#, but hopefully it should be fairly generic code. The important line is:

                        if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)

                        which strips the top i bits and XORs that with num right shifted. If the result is 0 then we have found an autocorrelation.

                        Jon CodeWrite

                        L Offline
                        L Offline
                        leppie
                        wrote on last edited by
                        #21

                        Very good! Wish I knew of it :(

                        IronScheme
                        ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                        J 1 Reply Last reply
                        0
                        • J Jonathan Nethercott

                          I think this should do it:

                          private int BinaryPeriod(int num)
                          {
                          int result = -1;
                          int numBits = (int)((Math.Log(num) / Math.Log(2))) + 1;

                          for (int i = 1; i <= numBits / 2 && result == -1; i++)
                          {
                              if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)
                                  result = i;
                          }
                          return result;
                          

                          }

                          I've done it in C#, but hopefully it should be fairly generic code. The important line is:

                          if (((num & ((1 << (numBits - i)) - 1)) ^ (num >> i)) == 0)

                          which strips the top i bits and XORs that with num right shifted. If the result is 0 then we have found an autocorrelation.

                          Jon CodeWrite

                          L Offline
                          L Offline
                          leppie
                          wrote on last edited by
                          #22

                          And here we have it in Scheme :) http://eval.ironscheme.net/?id=59[^] Thanks again!

                          IronScheme
                          ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                          1 Reply Last reply
                          0
                          • L leppie

                            Very good! Wish I knew of it :(

                            IronScheme
                            ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                            J Offline
                            J Offline
                            Jonathan Nethercott
                            wrote on last edited by
                            #23

                            Thank you :) I love solving problems like this - although that has distracted me a bit from what I'm supposed to be doing ;P Good luck with the job!

                            Jon CodeWrite

                            1 Reply Last reply
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                            • L leppie

                              Was given a problem that was suppose to come from the 'easy' bin. Two hours later I was still not finished... On the plus side, the interviewer attempted the problem at the same time, and also got stuck :) He said the problem must have been miscategorized. He told me to finish at home and to send the solution when done ;p Anyways, after a good nights sleep (with some wack apocalyptic dream I had), I started from scratch and coded a solution in about an hour this morning. The given problem: Binary period (dont bother googling for answers as you will likely just end up getting astronomy results or if you have google foo, a few very mathematical papers mostly related to cryptography) Given a number N larger than 0 (range was given, but does not matter), find the minimum period of repeating bit sequences or -1 if none is found. Examples

                              110110110 => 3 (as 110 repeats)
                              11011011 => 3 (still 3 and valid as the remaining subset, 11, is in 110)
                              110110111 => -1 (no period)
                              101 => -1 (no repeats, repeat implies at least 2 sequences)
                              10101 => 2 (repeat at least over 2 sequences)
                              11 => 1

                              10101010101 => 2
                              111011011101101110 => 7
                              111111111111111110 => -1

                              My answer in Scheme can be viewed and run here[^]. Edit: Web host seems to be down now... Edit 2: Seems to be back up again, please only look one at a time ;p Damn you shared hosting. If above link is down, view here[^].

                              IronScheme
                              ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                              E Offline
                              E Offline
                              Ennis Ray Lynch Jr
                              wrote on last edited by
                              #24

                              Heh, I had to do that in University.

                              Need custom software developed? I do custom programming based primarily on MS tools with an emphasis on C# development and consulting. I also do Android Programming as I find it a refreshing break from the MS. "And they, since they Were not the one dead, turned to their affairs" -- Robert Frost

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                              • E Ennis Ray Lynch Jr

                                Heh, I had to do that in University.

                                Need custom software developed? I do custom programming based primarily on MS tools with an emphasis on C# development and consulting. I also do Android Programming as I find it a refreshing break from the MS. "And they, since they Were not the one dead, turned to their affairs" -- Robert Frost

                                L Offline
                                L Offline
                                leppie
                                wrote on last edited by
                                #25

                                Was that some kind of homework or other task? Perhaps thesis? ;p

                                IronScheme
                                ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                                E 1 Reply Last reply
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                                • L leppie

                                  Was that some kind of homework or other task? Perhaps thesis? ;p

                                  IronScheme
                                  ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

                                  E Offline
                                  E Offline
                                  Ennis Ray Lynch Jr
                                  wrote on last edited by
                                  #26

                                  Well my degree was computer science. All we did was learn stuff like this. Wouldn't say I knew much about "programming" until after school. But solving problems with code, easy.

                                  Need custom software developed? I do custom programming based primarily on MS tools with an emphasis on C# development and consulting. I also do Android Programming as I find it a refreshing break from the MS. "And they, since they Were not the one dead, turned to their affairs" -- Robert Frost

                                  L 1 Reply Last reply
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                                  • E Ennis Ray Lynch Jr

                                    Well my degree was computer science. All we did was learn stuff like this. Wouldn't say I knew much about "programming" until after school. But solving problems with code, easy.

                                    Need custom software developed? I do custom programming based primarily on MS tools with an emphasis on C# development and consulting. I also do Android Programming as I find it a refreshing break from the MS. "And they, since they Were not the one dead, turned to their affairs" -- Robert Frost

                                    L Offline
                                    L Offline
                                    leppie
                                    wrote on last edited by
                                    #27

                                    Not even in my 2nd year course (just before I decided to go back to working class) we did stuff like this... I left when when we were busy with Data Structures and Algorithms.

                                    IronScheme
                                    ((λ (x) `(,x ',x)) '(λ (x) `(,x ',x)))

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